$$

Diagonalization

Problem

Let

$$A=\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right].$$

Find a basis

$\beta$ such that

$[f_A{]}_{\beta }^{\beta }$ is diagonal.

Thought

Let

$\beta =\{{\mathbf{u}}_1,{\mathbf{u}}_2\}$. Suppose

$$[f_A{]}_{\beta }^{\beta }=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right].$$

Then it means that

$$\begin{array}{rl}{f}_A({\mathbf{u}}_1)& =A{\mathbf{u}}_1={\lambda }_1{\mathbf{u}}_1,\\ f_A({\mathbf{u}}_2)& =A{\mathbf{u}}_2={\lambda }_2{\mathbf{u}}_2.\end{array}$$

Note that

$A\mathbf{v}=\lambda \mathbf{v}$ for some nonzero vector

$\mathbf{v}$ if and only if

$A-\lambda I$ is singular, which means

$det(A-\lambda I)=0$.

Sample answer

Compute

$$det(A-xI)=det\left[\begin{array}{cc}1-x& 2\\ 1& 2-x\end{array}\right]=x^2-3x=x(x-3),$$

which has zeros

$x=0,3$.

For

$x=3$, we have

$$A-3I=\left[\begin{array}{cc}-2& 2\\ 1& -1\end{array}\right],\text{ which as its reduced echelong form }\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right].$$

Thus, we may choose a nonzero vector

${\mathbf{u}}_1=\left[\begin{array}{c}1\\ 1\end{array}\right]$ in

$\ker (A-3I)$.

For

$x=0$, we have

$$A-0I=\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right],\text{ which as its reduced echelong form }\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right].$$

Thus, we may choose a nonzero vector

${\mathbf{u}}_2=\left[\begin{array}{c}2\\ -1\end{array}\right]$ in

$\ker (A-0I)$.

In summary, with

$\beta =\{{\mathbf{u}}_1,{\mathbf{u}}_2\}$, we have

$$[f_A{]}_{\beta }^{\beta }=\left[\begin{array}{cc}4& 0\\ 0& 6\end{array}\right],$$

so we found the desired basis

$\beta$.

Note

Please refer to How to solve a system of linear equations? if you need some help on how to find a basis of the kernel.

You may try other matrices such as

$$A=\left[\begin{array}{cc}5& -1\\ -1& 5\end{array}\right]$$

in quadratic curve.

This note can be found at Course website > Learning resources.